$$$\sqrt{x} \sqrt{1 - x}$$$ 的积分

求$$$\int \sqrt{x} \sqrt{1 - x}\, dx$$$。

输入已重写为：$$$\int{\sqrt{x} \sqrt{1 - x} d x}=\int{\sqrt{- x^{2} + x} d x}$$$。

配平方（步骤见»）：$$$- x^{2} + x = \frac{1}{4} - \left(x - \frac{1}{2}\right)^{2}$$$:

$${\color{red}{\int{\sqrt{- x^{2} + x} d x}}} = {\color{red}{\int{\sqrt{\frac{1}{4} - \left(x - \frac{1}{2}\right)^{2}} d x}}}$$

设$$$u=x - \frac{1}{2}$$$。

则$$$du=\left(x - \frac{1}{2}\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

所以，

$${\color{red}{\int{\sqrt{\frac{1}{4} - \left(x - \frac{1}{2}\right)^{2}} d x}}} = {\color{red}{\int{\sqrt{\frac{1}{4} - u^{2}} d u}}}$$

设$$$u=\frac{\sin{\left(v \right)}}{2}$$$。

则$$$du=\left(\frac{\sin{\left(v \right)}}{2}\right)^{\prime }dv = \frac{\cos{\left(v \right)}}{2} dv$$$（步骤见»）。

此外，可得$$$v=\operatorname{asin}{\left(2 u \right)}$$$。

因此，

$$$\sqrt{\frac{1}{4} - u ^{2}} = \sqrt{\frac{1}{4} - \frac{\sin^{2}{\left( v \right)}}{4}}$$$

利用恒等式 $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$：

$$$\sqrt{\frac{1}{4} - \frac{\sin^{2}{\left( v \right)}}{4}}=\frac{\sqrt{1 - \sin^{2}{\left( v \right)}}}{2}=\frac{\sqrt{\cos^{2}{\left( v \right)}}}{2}$$$

假设$$$\cos{\left( v \right)} \ge 0$$$，我们得到如下结果：

$$$\frac{\sqrt{\cos^{2}{\left( v \right)}}}{2} = \frac{\cos{\left( v \right)}}{2}$$$

积分变为

$${\color{red}{\int{\sqrt{\frac{1}{4} - u^{2}} d u}}} = {\color{red}{\int{\frac{\cos^{2}{\left(v \right)}}{4} d v}}}$$

对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(v \right)} = \cos^{2}{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$${\color{red}{\int{\frac{\cos^{2}{\left(v \right)}}{4} d v}}} = {\color{red}{\left(\frac{\int{\cos^{2}{\left(v \right)} d v}}{4}\right)}}$$

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，并令 $$$\alpha= v $$$:

$$\frac{{\color{red}{\int{\cos^{2}{\left(v \right)} d v}}}}{4} = \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 v \right)}}{2} + \frac{1}{2}\right)d v}}}}{4}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(v \right)} = \cos{\left(2 v \right)} + 1$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$\frac{{\color{red}{\int{\left(\frac{\cos{\left(2 v \right)}}{2} + \frac{1}{2}\right)d v}}}}{4} = \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 v \right)} + 1\right)d v}}{2}\right)}}}{4}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(\cos{\left(2 v \right)} + 1\right)d v}}}}{8} = \frac{{\color{red}{\left(\int{1 d v} + \int{\cos{\left(2 v \right)} d v}\right)}}}{8}$$

应用常数法则 $$$\int c\, dv = c v$$$，使用 $$$c=1$$$：

$$\frac{\int{\cos{\left(2 v \right)} d v}}{8} + \frac{{\color{red}{\int{1 d v}}}}{8} = \frac{\int{\cos{\left(2 v \right)} d v}}{8} + \frac{{\color{red}{v}}}{8}$$

设$$$w=2 v$$$。

则$$$dw=\left(2 v\right)^{\prime }dv = 2 dv$$$ (步骤见»)，并有$$$dv = \frac{dw}{2}$$$。

积分变为

$$\frac{v}{8} + \frac{{\color{red}{\int{\cos{\left(2 v \right)} d v}}}}{8} = \frac{v}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}}{8}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(w \right)} = \cos{\left(w \right)}$$$ 应用常数倍法则 $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$：

$$\frac{v}{8} + \frac{{\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}}{8} = \frac{v}{8} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{2}\right)}}}{8}$$

余弦函数的积分为 $$$\int{\cos{\left(w \right)} d w} = \sin{\left(w \right)}$$$：

$$\frac{v}{8} + \frac{{\color{red}{\int{\cos{\left(w \right)} d w}}}}{16} = \frac{v}{8} + \frac{{\color{red}{\sin{\left(w \right)}}}}{16}$$

回忆一下 $$$w=2 v$$$:

$$\frac{v}{8} + \frac{\sin{\left({\color{red}{w}} \right)}}{16} = \frac{v}{8} + \frac{\sin{\left({\color{red}{\left(2 v\right)}} \right)}}{16}$$

回忆一下 $$$v=\operatorname{asin}{\left(2 u \right)}$$$:

$$\frac{\sin{\left(2 {\color{red}{v}} \right)}}{16} + \frac{{\color{red}{v}}}{8} = \frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(2 u \right)}}} \right)}}{16} + \frac{{\color{red}{\operatorname{asin}{\left(2 u \right)}}}}{8}$$

回忆一下 $$$u=x - \frac{1}{2}$$$:

$$\frac{\sin{\left(2 \operatorname{asin}{\left(2 {\color{red}{u}} \right)} \right)}}{16} + \frac{\operatorname{asin}{\left(2 {\color{red}{u}} \right)}}{8} = \frac{\sin{\left(2 \operatorname{asin}{\left(2 {\color{red}{\left(x - \frac{1}{2}\right)}} \right)} \right)}}{16} + \frac{\operatorname{asin}{\left(2 {\color{red}{\left(x - \frac{1}{2}\right)}} \right)}}{8}$$

因此，

$$\int{\sqrt{- x^{2} + x} d x} = \frac{\sin{\left(2 \operatorname{asin}{\left(2 x - 1 \right)} \right)}}{16} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}$$

使用公式 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$，化简该表达式：

$$\int{\sqrt{- x^{2} + x} d x} = \frac{\sqrt{1 - \left(2 x - 1\right)^{2}} \left(2 x - 1\right)}{8} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}$$

加上积分常数：

$$\int{\sqrt{- x^{2} + x} d x} = \frac{\sqrt{1 - \left(2 x - 1\right)^{2}} \left(2 x - 1\right)}{8} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}+C$$

$$$\int \sqrt{x} \sqrt{1 - x}\, dx = \left(\frac{\sqrt{1 - \left(2 x - 1\right)^{2}} \left(2 x - 1\right)}{8} + \frac{\operatorname{asin}{\left(2 x - 1 \right)}}{8}\right) + C$$$A